Calculating Work Done by Force of Gravity on a 0.245 kg Ball

[alfredo24pr]

Homework Statement

A 0.245 kg ball is thrown straight up from 2.20 m above the ground. Its initial vertical speed is 12.40 m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.

Homework Equations

1/2m(v initial)2 + mg(h initial) = 1/2m(v final)2 +mg(h final)

The Attempt at a Solution

I know that y max equals v initial 2 /2g which is 7.84, but I do not know how to use the 2.2 which is the initial height

 

[alfredo24pr]

Bump
help

 

[Mr.A.Gibson]

A couple of ways to solve it I think, either
work = change in kinetic energy

or an easier way which is:
work = force x displacement (The displacement is 2.2m)

 

[alfredo24pr]

A couple of ways to solve it I think, either
work = change in kinetic energy

or an easier way which is:
work = force x displacement (The displacement is 2.2m)

I have no idea how to find the force. F=ma
How do I get a?

 

[Mr.A.Gibson]

I have no idea how to find the force. F=ma
How do I get a?

The force applied to the mass is gravity. Use F=ma, a is the acceleration due to gravity g=9.81m/s^2. Sometimes rounded to 10m/s^2

The displacement is 2.2m since this is the distance moved from start to finish.

Work = Mass x acceleration x displacement
= 0.245 x 9.81 x 2.2

 

[alfredo24pr]

The force applied to the mass is gravity. Use F=ma, a is the acceleration due to gravity g=9.81m/s^2. Sometimes rounded to 10m/s^2

The displacement is 2.2m since this is the distance moved from start to finish.

Work = Mass x acceleration x displacement
= 0.245 x 9.81 x 2.2

That is mgh which is potential energy. The ball is thrown upward at 12.4 m/s

 

[Mr.A.Gibson]

That is mgh which is potential energy. The ball is thrown upward at 12.4 m/s

You are correct gravitational potential energy GPE is another way of stating the work equation, i.e. they are the same.

The ball is throw upward at 12.4m/s, so if there is no air resistance then it will be traveling downwards past the same point also at -12.4m/s. So on the way up and the way down at this point the energy calculation is the same.
$$E = \frac{1}{2} m v^2 + m g h$$

So the total workdone by gravity on the balls path up and down to the same point is zero. This is because it travels in the opposite direction with the same force acting on it.

So the only part we need to calculate is the potential energy lost as it falls the final 2.2m to the ground. This is the total work done by gravity.

Another way to approach it would be that we only look at the energy and the start when it is thrown and the energy at the end when it hits the ground. The energy transferred from GPE to KE in this situation is then the work done by gravity.

Hope this is clear

 

[alfredo24pr]

E= 1/2mv2 + mgh
E= 1/2 (0.245)(12.4)2 + 0.245*9.8*2.2
E= 18.8356 + 5.2822
E= 24.1178 J

but isn't that only the initial
I need Ki + Ui = Kf + Uf
I have Ki + Ui = 24.1 J, what am i missing?

 

[madah12]

Even thought E1=E2 is correct it isn't the easiest way to solve this ,but delta k =W is a better way.
You should calculate the final velocity to get Kf and then get the work

 

[alfredo24pr]

Even thought E1=E2 is correct it isn't the easiest way to solve this ,but delta k =W is a better way.
You should calculate the final velocity to get Kf and then get the work

How do I use the delta K way? please help.

 

[alfredo24pr]

BUMP
help meee!

 

[alfredo24pr]

double bump
HELP!